A school wants to award its students for the value of honesty, regularity and hard work will total cash award of Rs. `6000.` Three times the award money for hard work added to that added to that given for honesty amounts to Rs. `11000.` The award money given for honesty and hard work together is double the one given for regularity. Represent the above situation algebraically and find the award money for each value, using matrix method.

Explanation

Let the amount of award for Honesty . Regularly and Hardwork be ₹x,₹y,and ₹z respectively . Then
`x+y+z=6000, " " ...(i)`
`x+0y+3z=11000 " " ...(i)`
` and x+z=2y rArr x-2y+z=0 " " ...(iii)`
Let `A=[{:(1" "1" "1),(1" "0" "3),(1" "-2" "1)] ,X=[(x),(y),(z)] and B=[(600),(11000),(0)]:}`
Then,the matrix equation is Ax =B
`therefore X=A^(-1)B`
Now, `|A|={:[(1" "1" "1),(1" "0" "3),(1""-2" "1)]=[(1" "0" "0),(1-1" "2),(1-3" "0)]:}`
`[C_2 rarr (C_2-C_1) and (C_3 rarr C_1)]`
`= 1.|(-1" "2),(-3" "0)|=(0+6)=6 ne 0`
`thererfore A^(-1)` exists
Now the cofactors of the elements of |A| are given by
`C_(11)=|(0,3),(-2,1)|=6,C_12=-|(1,3),(1,1)|=2,C_13=|(1,0),(1,-2)|=-2`
`C_21=-|(1,1),(-2,1)|=-3,C_22=|(1,1),(1,1)|=0,C_23=-|(1,1),(1,-2)|=3`
`C_31=|(1,1),(0,3)|=3,C_32=-|(1,1),(1,3)|=-2,C_33=|(1,1),(1,0)|=-1`
`therefore (adj A)=[(6,2,-2),(-3,0,3),(3,-2,-1)]^7=[(6,-3,3),(2,0,-2),(-2,3,-1)]`
`rArr A^(-1)=1/(|A|),(adj A)=1/6.[(6,-3,3),(2,0,-2),(-2,3,-1)]`
`rArr X-A^(-1)B=1/6.[(6,-3,3),(2,0,2),(-2,3,-1)]" "[(6000),(11000),(0)]`
`rArr =1/6[(3600-33000+0),(12000+" "0" "+0),(-12000+33000-0)]=1/6[(3000),(12000),(21000)]`
`rArr [(x),(y),(z)]=[(500),(2000),(3500)]`
`rArr x=500,y=2000 and z=3500`
Hence, the award money for Honesty,Regularity and Hardwork is ₹ 500 , ₹ 2000 and ₹ 3500 respectively .
Apart from hoenesty and hardwork the school must include an award for a student to be well- behaved .